Question

b)a physical quantity p is related to four different a,b,c and d as p = a^3 b^2/root cd. then percentage error in measurement of a,b,c and d are 1%,3%,2% and 3% respectively.what is the percentage error in the measurement of p?​

Answer 1

Answer:

11.5

Explanation:

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Answer 2

Answer:

• percentage error in p = 13%

Explanation:

Given that,

Physical quantity p is related to four quantities a, b, c and d as

$\rm{p = \dfrac{\;\;a^3 b^2}{\sqrt{c\;}\;d\;}}$

• Percentage errors in measurements of a, b, c and d are 1%, 3%, 2% and 3% respectively.

We need to find,

• Percentage error in measurement of p

Solution,

Since, percentage errors in a, b, c and d are 1%, 3%, 2% and 3% respectively

therefore,

$\rm{\dfrac{\Delta a}{a}\times 100\% = 1\% \implies \dfrac{\Delta a}{a} = 0.01}$

$\rm{\dfrac{\Delta b}{b}\times 100\% = 3\% \implies \dfrac{\Delta b}{b} = 0.03}$

$\rm{\dfrac{\Delta c}{c}\times 100\% = 2\% \implies \dfrac{\Delta c}{c} = 0.02}$

$\rm{\dfrac{\Delta d}{d}\times 100\% = 3\% \implies \dfrac{\Delta d}{d} = 0.03}$

Given relation is

$\rm{p = \dfrac{\;\;a^3 b^2}{\sqrt{c\;}\;d\;}}$

so,

$\rm{\dfrac{\Delta p}{p} = 3\dfrac{\Delta a}{a} + 2\dfrac{\Delta b}{b} + \dfrac{1}{2}\dfrac{\Delta c}{c} + \dfrac{\Delta d}{d}}$

$\rm{\dfrac{\Delta p}{p} = 3 \times (0.01) + 2\times (0.03) + \dfrac{1}{2} \times (0.02) + (0.03)}$

$\rm{\dfrac{\Delta p}{p} = 0.03 + 0.06 +0.01 + 0.03}$

$\rm{\dfrac{\Delta p}{p} = 0.13}$

Calculating the percentage error in p

$\rm{\dfrac{\Delta p}{p} \times 100\% = 0.13 \times 100\% = 13\%}$

Therefore,

• The percentage error in quantity p is 13%.